How many grams of al(oh)3 (molar mass = 78.0 g/mol) can be produced from the reaction of 48.6 ml of .15 m koh with excess al2(so

Question

bagirrra123 [75]

2 years ago

9

How many grams of al(oh)3 (molar mass = 78.0 g/mol) can be produced from the reaction of 48.6 ml of .15 m koh with excess al2(so

4)3?

Chemistry

1 answer:

Fudgin [204] · Answer 1 · 2023-07-21T08:45:55+03:00

Taking into account the reaction stoichiometry, 0.18954 grams of Al(OH)₃ are formed from the reaction of 48.6 mL (0.0486 L) of 0.15 M KOH with excess Al₂(SO₄)₃.

<h3>Reaction stoichiometry</h3>

In first place, the balanced reaction is:

Al₂(SO₄)₃ + 6 KOH → 2 Al(OH)₃ + 3 K₂SO₄

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

Al₂(SO₄)₃: 1 mole
KOH: 6 moles
Al(OH)₃: 2 moles
K₂SO₄: 3 moles

The molar mass of the compounds is:

Al₂(SO₄)₃: 342 g/mole
KOH: 56.1 g/mole
Al(OH)₃: 78 g/mole
K₂SO₄: 174.2 g/mole

Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

Al₂(SO₄)₃: 1 mole ×342 g/mole= 342 grams
KOH: 6 moles ×56.1 g/mole= 336.6 grams
Al(OH)₃: 2 moles ×78 g/mole= 156 grams
K₂SO₄: 3 moles ×174.2 g/mole= 522.6 grams

<h3>Definition of molarity</h3>

Molarity is a measure of the concentration of a solute in a solution and indicates the number of moles of solute that are dissolved in a given volume:

Molarity= number of moles÷ volume

<h3>Mass of Al(OH)₃ formed</h3>

In firts place, you know that 48.6 mL (0.0486 L) of 0.15 M KOH react with excess Al₂(SO₄)₃.

Replacing in the definition, you can calculate the amount of moles of KOH that react:

0.15 M= number of moles÷ 0.0486 L

Solving:

0.15 M× 0.0486 L= number of moles

<u><em>0.00729 moles= number of moles</em></u>

Then, 0.00729 moles of KOH react with excess Al₂(SO₄)₃. So, following rule of three can be applied: if by reaction stoichiometry 6 moles of KOH form 156 grams of Al(OH)₃, 0.00729 moles of KOH form how much mass of Na₂SO₄?

$mass of Al(OH)_{3} =\frac{0.00729 moles of KOHx156 grams of Al(OH)_{3}}{6 moles of KOH}$